A cord is wound over the rim of a flywheel of mass 20 kg and radius 25 cm. A mass 2.5 kg attached to the cord is allowed to fall under gravity. The angular acceleration of the flywheel is

Options

A. `25 rad//s^(2)`
B. `20 rad//s^(2)`
C. `10 rad//s^(2)`
D. `5 rad//s^(2)`

Answer

C

Explanation

Moment of inertia of flywheel about its axis ,
`I = (MR^(2))/(2) = ((20 kg) (25 xx 10^(-2) m)^(2))/(2)`
`= 25 xx 25 xx 10^(-3) kg m^(2)`
Torque acting on the flywheel , `tau = FR = mg R `
`= (2.5 kg) (10 m//s^(2)) (25 xx 10^(-2) m)`
`=25 xx 25 xx 10^(-2) Nm`
Angular acceleration of the flywheel ,
`alpha = (tau)/(I) = (25 xx 25 xx 10^(-2))/(25 xx 25 xx 10^(-3)) = 10 "rad/s"^(2)`